# Limit with arctan: $\lim_{x\rightarrow 0} \frac{x\sin 3x}{\arctan x^2}$

$$\lim_{x\rightarrow 0} \frac{x\sin 3x}{\arctan x^2}$$

NB! I haven't learnt about L'Hôpital's rule yet, so I'm still solving limits using common limits.

### What I've done so far

$$\lim_{x\rightarrow0}\left[\frac{x}{\arctan x^2}\cdot 3x\cdot \frac{\sin 3x}{3x}\right] = 0\cdot1\cdot\lim_{x\rightarrow0}\left[\frac{x}{\arctan x^2}\right] = 0??$$ Obviously I'm wrong, but I thought I'd show what I tried.

• Do you have this inequality: $\sin x \leq x \leq \tan x$ for $x \geq 0$? And the limits that follow from it: $\sin x / x \rightarrow 0$, etc.? – Simon S Nov 3 '14 at 23:46

## 4 Answers

You should put in $$\frac{x^2}{\arctan x^2}$$ that has limit $1$: $$\lim_{x\to0}\frac{x\sin3x}{\arctan x^2}= \lim_{x\to0}3\frac{\sin3x}{3x}\frac{x^2}{\arctan x^2}=\dots$$ If you don't know the limit above, just substitute $t=\arctan x^2$, so $x^2=\tan t$ and the limit is $$\lim_{x\to0}\frac{x^2}{\arctan x^2}=\lim_{t\to0}\frac{\tan t}{t}$$ that you should be able to manage.

Starting as you did, we want to find the limit of $$\frac{3x^2}{\arctan(x^2)}\cdot\frac{\sin(3x)}{3x}.$$ Only the first term gives any trouble. Let $x$ be not too large, and let $x^2=\tan w$, You want to find $$\lim_{w\to 0} \frac{3\tan w}{w},$$ which is not difficult. Replace $\tan w$ by $\frac{\sin w}{cos w}$.

Since $u \sim \tan u$, then $y \sim \arctan y$, so $x^2 \sim \arctan (x^2)$.

This means that $$\lim_{x \to 0} \frac{x \sin 3x}{\arctan x^2} = \lim_{x \to 0} \frac{x3x}{x^2} = 3$$

• This feels unsatisfactory, as you are still using the derivatives of the functions in the numerator and denominator. – Simon S Nov 3 '14 at 23:42

Using the Asymptotic expansion:
$\sin(x)\approx_0 x$
$\arctan(x^2) \approx_0 x^2$
So $$\lim _{x\rightarrow \:0}\:\frac{x\sin \:3x}{\arctan \:x^2}=\lim \:_{x\rightarrow 0}\:\frac{3x^2}{x^2}=\color{red}{3}$$